Problem 1 (10 points) The present problem focuses on calculating the solutions,

Question

Problem 1 (10 points) The present problem focuses on calculating the solutions, (x,y, z) defined by the following augmented matrix [A, , if any, of the linear system A, b Include at least the following information in your solution: (I.1) (6 points) Bring into reduced echelon form the augmented matrix (A, b]. Show every arithmetic operation. (1.2) (3 point) Calculate all the solutions,(z.y, 2), if any exist. Show every arithmetic operation. (1.3) (1 point) Verify your result by substituting your solutions into the system. Show every arithmetic operation. Optionally check all your results with any computing system. Note any discrepancy, if any arises.

Explanation / Answer

a)

given agumented matrix is

1.0000 + 1.0000i 1.0000 - 1.0000i -1.0000 + 1.0000i 1.0000   
-1.0000 + 3.0000i 4.0000 + 1.0000i -3.0000 - 1.0000i 0   
-1.0000 + 5.0000i 7.0000 + 3.0000i -7.0000 - 3.0000i 0

R1-> R1/(1+i)

1.0000 0 - 1.0000i 0 + 1.0000i 0.5000 - 0.5000i
-1.0000 + 3.0000i 4.0000 + 1.0000i -3.0000 - 1.0000i 0   
-1.0000 + 5.0000i 7.0000 + 3.0000i -7.0000 - 3.0000i 0   

R2->R2-R1*(-1+3i)

1.0000 1.0000i 1.0000i 0.5000 - 0.5000i
0 1.0000 0 -1.0000 - 2.0000i
-1.0000 + 5.0000i 7.0000 + 3.0000i -7.0000 - 3.0000i 0   

R3->R3-R1*(-1+5i)

1.0000 - 1.0000i 0 + 1.0000i 0.5000 - 0.5000i
0 1.0000 0 -1.0000 - 2.0000i
0 2.0000 + 2.0000i -2.0000 - 2.0000i -2.0000 - 3.0000i

R1->R1+R2*i

1.0000 0 0 + 1.0000i 2.5000 - 1.5000i
0 1.0000 0 -1.0000 - 2.0000i
0 2.0000 + 2.0000i -2.0000 - 2.0000i -2.0000 - 3.0000i

R3-> R3-R2*(2+2i)

1.0000 0 0 + 1.0000i 2.5000 - 1.5000i
0 1.0000 0 - 1.0000 - 2.0000i
0 0 -2.0000 - 2.0000i -4.0000 + 3.0000i

R3->R3/(-2-2i)

1.0000 0 0 + 1.0000i 2.5000 - 1.5000i
0 1.0000 0 -1.0000 - 2.0000i
0 0 1.0000 0.2500 - 1.7500i

R1->R1-R3*i

1.0000 0 0 0.7500 - 1.7500i
0 1.0000 0 -1.0000 - 2.0000i
0 0 1.0000 0.2500 - 1.7500i

b)

thus

x=  0.7500 - 1.7500i  

y= -1.0000 - 2.0000i

z=0.2500 - 1.7500i

c)

checking the solution

(1+i)x -(1-i)y+(i-1)*z= 1

LHS= (1+i)*(0.75-1.75i)+(1-i)*(-1-2*i)+(i-1)*(0.25-1.75*i)

=2.5-i -3 -i +1.5+2i

=1

=RHS

Simillerly we can satisfy other solution too

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.